Aaron is reading a graph of equivalent ratios. If one of the ratios is represented as (6, 2), which could be another ratio?

2 answers:

6 0

There Are Unlimited Equivalent Ratios To This! For One, We Could Simplify This Ratio To 3,1.Another One Would Be 12,4. You See, It Can Be Any Ratio As Long As It Equals(Simplified) To 3/1. I Hope This Helps!

MaRussiya [10]3 years ago

5 0

Answer:

(18, 6)

Step-by-step explanation:

Equivalent ratios are ratios that are the same when reduced ro lowest expressions. They can be generated by multiplying a given ratio by consecutive numbers, not simultaneously.

So for this question, we can generate another ratio by multiplying through by 3,

= (6×3, 2×3)

= (18, 6)

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4 years ago

Solve for x: 4 over x plus 4 over quantity x squared minus 9 equals 3 over quantity x minus 3. (2 points) Select one: a. x = -4

Kay [80]

Answer:

Step-by-step explanation:

$\dfrac{4}{x}+\dfrac{4}{x^2-9}=\dfrac{3}{x-3}$

Domain:

$x\neq0\ \wedge\ x^2-9\neq0\ \wedge\ x-3\neq0\\\\x\neq0\ \wedge\ x\neq\pm3$

solution:

$\dfrac{4}{x}+\dfrac{4}{x^2-3^2}=\dfrac{3}{x-3}$

use <em>(a - b)(a + b) = a² - b²</em>

$\dfrac{4}{x}+\dfrac{4}{(x-3)(x+3)}=\dfrac{3}{x-3}$

multiply both sides by (x - 3) ≠ 0

$\dfrac{4(x-3)}{x}+\dfrac{4(x-3)}{(x-3)(x+3)}=\dfrac{3(x-3)}{x-3}$

cancel (x - 3)

$\dfrac{4(x-3)}{x}+\dfrac{4}{x+3}=3$

subtract $\frac{4(x-3)}{x}$ from both sides

$\dfrac{4}{x+3}=3-\dfrac{4(x-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x}{x}-\dfrac{(4)(x)+(4)(-3)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-\bigg(4x-12\bigg)}{x}\\\\\dfrac{4}{x+3}=\dfrac{3x-4x-(-12)}{x}\\\\\dfrac{4}{x+3}=\dfrac{-x+12}{x}$